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4 years ago

If a triangle is a right triangle, then the other two angles must be acute. true or false

1 answer:

5 0

That depends on which angle you picked first, because that determines
which angles "the other two" are.

If you picked the right angle (90°) first, before you asked the question,
then the other two are acute angles, and they're also complementary

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2. What value did you calculate for the index of refraction of the acrylic block in Part 2? How does your value compare to the a

lesya [120]

The percentage error in calculated value of the refractive index of acrylic block is calculated using the formula: (true value - calculated value/true value) × 100%.

<h3>What is refractive index of a refracting medium?</h3>

The refractive index of a medium is the ratio of the sine of the angle of incidence on the medium and the sine of the angle of refraction.

Refractive index = sin i/sin r

where;

i is the angle of incidence
r is angle of refraction.

The refractive index of a medium increases with density of the medium.

To determine the percentage error in determining the refractive index of a medium, the following formula is used:

Percent error = error/true value × 100%

Therefore, the percentage error in calculated value of the refractive index of acrylic block is the ratio of the error and true value multiplied by 100%.

Learn more about refractive index at: brainly.com/question/83184

8 0

3 years ago

Learning Goal: To review the concept of conservative forces and to understand that electrostatic forces are, in fact, conservati

CaHeK987 [17]

Explanation:

The electrostatic forces are conservative forces!

The mainly property of the conservative fields is $\vec{\nabla} \times \vec E=\vec 0$

In spherical coordinates the field's expression is:

$\vec E=\frac{Q}{4\pi \epsilon _0 r^2} .\^r$

and the curl expression is:

$\nabla\times \vec E=\frac{1}{r^2{\sin}\,\theta}\left|\begin{matrix}\hat{r} & r\,\hat{\theta} & r\,{\sin}\,\theta\,\hat{\varphi} \\& & \\\frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \varphi}\\ & & \\E_r & rE_\theta & r{\sin}\,\theta\, E_\varphi\end{matrix}\right|=(0, 0, 0)$

to find the expression for the potential function associated:

$\vec E=\vec \nabla . V, \Delta V= V_b-V_a=-\int _c \vec E.d\vec l=-\int _c E\^r.dr\^r=-\int _c Edr=\int \limits^a_b \frac{Q}{4\pi \epsilon _0 r^2} dr= \frac{Q}{4\pi \epsilon _0}.(\frac{1}{r}|^b_a)= \frac{Q}{4\pi \epsilon _0}.(\frac{1}{b}-\frac{1}{a})$

5 0

3 years ago

Please help! this this timed!!!

VMariaS [17]

Answer:

Thermal heat

Explanation:

The water heats up which it means it is thermal (heat/hot)

4 0

3 years ago

Read 2 more answers

What is the critical angle for the interface between water and light flint? nflint= 1.58, nwater=1.33?

Angelina_Jolie [31]

When light crosses the interface between a medium with higher refractive index and a medium with lower refractive index, there is a maximum value of the angle of incidence after which there is no refraction, but all the light is reflected, and this maximum value is called critical angle.

The critical angle is given by
$\theta_C = \arcsin ( \frac{n_2}{n_1} )$
where n1 is the refractive index of the first medium while n2 is the refractive index of the second medium. In our problem, n1=1.33 and n2=1.58, so the critical angle is
$\theta_C = \arcsin ( \frac{1.33}{1.58} )=57.3 ^{\circ}$

3 0

3 years ago

A shot putter shoots a 7.3kg shot from rest to 14m/s in 1.5s. what was the average power?

sweet [91]

Answer: 477W

Explanation:

Given the following :

Mass (m) = 7.3kg

Initial Velocity (u) = 0

Final velocity (v) = 14m/s

time (t) = 1.5s

Power = workdone (W) / time (t)

The workdone can be calculated as the change in kinetic energy (KE) :

Recall ;

KE = 0.5mv^2

Therefore, change in KE is given by:

0.5mv^2 - 0.5mu^2

Change in KE = 0.5(7.3)(14^2) - 0.5(7.3)(0^2)

Change in KE = 715.4J

Therefore ;

Average power = Workdone / time

Workdone = change in KE = 715.4N

Average power = 715.4 / 1.5

Average power = 476.93333 W

= 477W

5 0

4 years ago